Problems on Distance Between Two Points
Solving the problems on distance between two points with the help of the formula, in the below examples use the formula to find distance between two points.
Worked-out problems on distance between two points:
1. Show that the points (3, 0), (6, 4) and (- 1, 3 ) are the vertices of a right-angled Isosceles triangle.
Solution: Let the given points be A(3, 0), B (6, 4) and C (-1, 3). Then we have,
AB^{2} = (6 - 3)^{2} + (4 - 0)^{2} = 9 + 16 = 25;
BC^{2} = (-1 - 6)^{2} + (3 - 4 )^{2} = 49 + 1= 50
and CA^{2} = (3 + 1)^{2} + (0 - 3)^{2} = 16 + 9= 25.
From the above results we get,
AB^{2} = CA^{2} i.e., AB = CA,
which proves that the triangle ABC is isosceles.
Again, AB^{2} + AC^{2} = 25 + 25 = 50 = BC^{2}
which shows that the triangle ABC is right-angled.
Therefore, the triangle formed by joining the given points is a right-angled isosceles triangle. Proved.
2. If the three points (a, b), (a + k cos α, b + k sin α) and (a + k cos β, b + k sin β) are the vertices of an equilateral triangle, then which of the following is true and why ?
(i) | α - β| = π/4
(ii) |α - β| = π/2
(iii) |α - β| = π/6
(iv) |α - β| = π/3
Solution: Let the vertices of the triangle be A (a, b), B (a + k cos α, b + k sin α) and C (a + k cos β, b + k sin β).
Now, AB^{2} = (a + k cos α - a)^{2} + (b + k sin α - b)^{2}
= k^{2} cos^{2} α + k^{2} sin^{2} α = k^{2};
Similarly, CA^{2} = k^{2} and
BC^{2} = (a + k cos β - a - k cos α)^{2} + (b + k sin β - b - k sin α)^{2}
= k^{2} (cos^{2} β + cos^{2} α - 2 cos α cos β + sin^{2} β + sin^{2} α - 2 sin α sin β)
= k^{2} [cos^{2} β + sin^{2} β + cos^{2} α + sin^{2} α - 2(cos α cos β + sin α sin β)]
= k^{2} [1 + 1 - 2 cos (α - β)]
= 2k^{2} [1 - cos (α - β)]
Since ABC is an equilateral triangle, hence
AB^{2} = BC^{2}
or, k^{2} = 2k^{2} [1 - cos (α - β)]
or, 1/2 = 1 - cos(α - β) [since, k # 0]
or, cos (α - β) = 1/2 = cos π/3
Therefore, |α - β| = π/3 .
There for, condition (iv) is true.
3. Find the point on the y-axis which is equidistant from the points (2, 3)and(-1, 2).
Solution: Let P(0, y) be the required point on the y-axis and the given points are A (2, 3) and B(- 1, 2). By question,
PA = PB = PA^{2} = PB^{2}
or, (2 - 0)^{2} + (3 - y)^{2} = (-1 - 0)^{2} + (2 – y)^{2}
or, 4 + 9 + y^{2} - 6y = 1 + 4 + y^{2} - 4y
or, - 6y + 4y = 1 - 9 or, - 2y = -8
or, y = 4.
Therefore, the required point on the y-axis is (0, 4).
4. Find the circum-centre and circum-radius of the triangle whose vertices are (3, 4), (3, - 6) and (- 1, 2).
Solution: Let A(3, 4), B (3, - 6), C (- 1, 2) be the vertices of the triangle and P(x, y ) the required circum-centre and r the circum-radius. Then, we must have,
r^{2} = PA^{2} = (x - 3)^{2} + (y - 4)^{2} ……………………..(1)
r^{2} = PB^{2} = (x - 3)^{2} + (y + 6)^{2} ……………………….(2)
and r^{2} = PC^{2} = (x + 1)^{2} + (y - 2)^{2} ……………………….(3)
From (1) and (2) we get,
(x - 3)^{2} + (y - 4)^{2} = (x - 3)^{2} + (y + 6)^{2}
Or, y^{2} - 8y + 16 = y^{2} + 12y + 36
or, - 20y = 20 or, y = - 1
Again, from (2) and (3) we get,
(x - 3)^{2} + (y + 6)^{2} = (x + 1 )^{2} + (y - 2)^{2}
or, x^{2} - 6x + 9 + 25 = x^{2} + 2x + 1 + 9 [putting y = - 1]
or, - 8x = - 24
or, x = 3
Finally, putting x = 3 and y = - 1 in (1) we get,
r^{2} = 0^{2} + (-1 - 4)^{2} = 25
Therefore, r = 5
Therefore, the co-ordinates of circum-centre are (3, - 1) and circum-radius = 5 units.
5. Show that the four points (2, 5), (5, 9), (9, 12) and (6, 8) when joined in order, form a rhombus.
Solution: Let the given points be A(2, 5), B (5, 9), C (9, 12) and D(6, 8). Now,AB^{2} = (5 - 2)^{2} + (9 - 5)^{2} = 9 + 16 = 25
BC^{2} = (9 - 5)^{2} + (12 - 9)^{2} = 16 + 9 = 25
CD^{2} = (6 - 9)^{2} (8 - 12)^{2} = 9 + 16 = 25
DA^{2} = (2 - 6)^{2} + (5 - 8)^{2} = 16 + 9 = 25
AC^{2} = ( 9 - 2)^{2} + (12 - 5)^{2} = 49 + 49 = 98
and BD^{2} = (6 - 5)^{2} + (8 - 9)^{2} = 1 + 1 = 2
From the above result we see that
AB = BC = CD = DA and AC ≠ BD.
That is the four sides of the quadrilateral ABCD are equal but diagonals AC and BD are not equal. Therefore, the quadrilateral ABCD is a rhombus. Proved.
The above worked-out problems on distance between two points are explained step-by-step with the help of the formula.
● Co-ordinate Geometry
What is Co-ordinate Geometry?
Rectangular Cartesian Co-ordinates
Polar Co-ordinates
Relation between Cartesian and Polar Co-Ordinates
Distance between Two given Points
Distance between Two Points in Polar Co-ordinates
Division of Line Segment: Internal & External
Area of the Triangle Formed by Three co-ordinate Points
Condition of Collinearity of Three Points
Medians of a Triangle are Concurrent
Apollonius' Theorem
Quadrilateral form a Parallelogram
Problems on Distance Between Two Points
Area of a Triangle Given 3 Points
Worksheet on Quadrants
Worksheet on Rectangular – Polar Conversion
Worksheet on Line-Segment Joining the Points
Worksheet on Distance Between Two Points
Worksheet on Distance Between the Polar Co-ordinates
Worksheet on Finding Mid-Point
Worksheet on Division of Line-Segment
Worksheet on Centroid of a Triangle
Worksheet on Area of Co-ordinate Triangle
Worksheet on Collinear Triangle
Worksheet on Area of Polygon
Worksheet on Cartesian Triangle
11 and 12 Grade Math
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